AcWing 2. 01背包问题

朴素

//朴素二维
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[N][N];
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
        cin >> v[i] >> w[i];
    for(int i=1; i<=n; i++)//装的个数 
        for(int j=0; j<=m; j++) {//最大容量 
        //讲 f[i][j]分为f[i-1][j](去掉第i个)和 f[i-1][j-v[i]]+w[i](先去掉第i个,并减去他的质量,再加上) 
            f[i][j]=f[i-1][j];//左边
            //右边 
            //当j<v[i]时,情况不存在,就不用考虑 
            if(j>=v[i]) f[i][j]=max(f[i][j],f[i-1][j-v[i]]+w[i]);
        }
    cout<<f[n][m]<<endl;
    return 0;
}

优化

//一维优化
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[N];
int main() {
    cin >> m >> n;//m表示时间,n表示数量
    for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];
    for (int i = 1; i <= n; i ++ )
        for (int j = m; j >= v[i]; j -- )
            f[j] = max(f[j], f[j - v[i]] + w[i]);
    cout << f[m];
    return 0;
}